\subsection*{2012-06-25}

\begin{enumerate}
  \item Computation of the AdS metric in higher dimension using Taylor series. 
  The assumed form of the metric was $g = \tfrac{1}{z^2}\,\mathrm{diag}\left(f(z),\, r(z)^2, r(z)^2 \sin^2(x_1), q(z)\right)$.
  The expansion equaqtions looked formidable at first. 
  The first degree of freedom beyond the boundary condiditons in $d+1$ AdS space is introduced at order $d$. 
  The terms of order less than $d$ that appear in the expansion are the same for different $d$'s. 
  We can kill all of those terms by an appropriate choice of $q(z)$. 
  In particular, computing in a high $d$ case, it looks like  $q(z)$ is most naturally $q(z) = \tfrac{1}{1 - z^2/r(0)}$.
  For that choice of $q$, the metric looks like: 
  \begin{equation}
  \begin{split}
    f(z) &= f(0) + \tfrac{a}{3!}z^3 + O(z^4) \\
    r(z) &= r(0) - \tfrac{1}{2!\,r(0)}z^2 - \tfrac{a}{4!} \tfrac{r(0)}{f(0)}z^3 + O(z^4)
    \end{split}
  \end{equation}
  The higher order terms get messier and messier; in particular, the series does not terminate.

  As it turns out, the series expansion in $z$ is not the natural thing to do. 
  In setting up the calculation we took a solution from $2+1$ case and extrapolated to higher dimension. 
  Unfortunately, the differential equations from $G_{\mu \nu} = 0$ that govern the form of $f(z)$ and $r(z)$ change as $d$ changes, and in particular, for $d = 2$ simplify as $(d-2)=0$.
  So, in the absence of a reasonable guess to start with, it's much easier to solve the differential equations directly. 
  Also, we have used up all of gauge power and symmetries to simplify the metric, so it is not unreasonable to expect that the differential equations are actually not too bad. 
  In finding the differential equations solution, we need to additionally impose that the asymptotic form of the solution is what we want. 
  Additionally, if we compute the curvature $R$ we discover $R = -12 +\left(6-\tfrac{4}{r(0)^2}+\tfrac{2}{r(0)}\right)z^2 + \tfrac{a}{4 f(0)} z^3$. 
  As it turns out, the $z^2$ vanishes in the full analytic solution. How can that be? The value of $r(0)$ is actually constrained by the EFE.
  The $z^3$ term is genuine, and represents curvature contribution from a black hole, whose mass is controlled by the parameter $a$.


  \item
  Origin of the Hawking boundary term in the GR action. 
  We want to obtain the EFE through varying of an action. 
  \begin{equation}
    S_{EH} = \tfrac{1}{2 \kappa} \int_M d^{d+1}x \sqrt{-g} R
  \end{equation}
  However, we end up with the bulk local term (EFE) \emph{and} a boundary term. 
  It looks like the Hawking boundary action arises just to offset this extra contribution, but turns out the story is a little more subtle.

  We first discuss the case of a scalar field to illustrate what is it we're after.  
  Take
  $$S = \int d^D x\, \left(\partial \phi\right)^2$$
  After varying the action we obtain 
  $$\int d^D x\, \partial^2 \phi \, \delta \phi  + \int d^Dx\, \phi \, \partial \phi\, \delta \phi = 0 $$
  That clearly does not lead to the familiar local $\partial^2 \phi = 0$.
  We need to additionally require that $\delta \phi = 0$ at the boundary, in which case the second integral vanishes. 
  Why can we do this? 
  Think about how local perturbations of the field affect the boundary value of the field. 
  If the area of the boundary grows sufficiently fast away from the point of perturbation, then the field perturbation dilutes as it propagates outwards, and the amount of energy required to change the boundary field value would need to be infinite.  

  Think of an analogy with quantum mechanics: If you start out with a gaussian, it spreads out as time goes on, and the value of the boundary remains unchanged.
  A field configuration with non-zero boundary value may or may not result in a finite action. 
  If the theory does not admit the possibility of changing the boundary value with a finite perturbation (amount of energy), then solutions with non-zero boundary values will necessarily be non-normalizable -- the corresponding action will diverge. 

  Now, let's leave the question of normalizability for later and focus on the boundary term in the action. 
  If the theory admits normalizable solutions with non-zero boundary values, we need to worry about the following situation.
  If we have a solution $\phi$ with zero boundary values, we may construct a different solution $\tilde{\phi} = \phi + \delta \phi$, where $\delta \phi$ is some solution of the EOM that are mostly zero in the bulk but non-zero towards the boundary.
  We would like $\tilde{\phi}$ to be an extremum of the action as well. 
  However, if we do not offset the boundary term in the variation of action, the equations of motion for the two cases are different, and so $\tilde{\phi}$ is not a solution of $\partial^2 \phi$!
  Hence, we're motivated to introduce another boundary term into the action by hand. 

  Can we ignore the boundary term (analoguous to setting $\delta g = 0$ at the boundary) in the gravity action? 
  In principle -- no! 
  The metric is the dynamical quantity appearing in the action, and physically we expect the metric to respond to appropriately chosen initial conditions everywhere, and in particular at the boundary -- throw an apple and it'll eventually reach the boundary. 
  Most generally, it may even happen that a combination of initial conditions and boundary conditions are \emph{incompatible} with one another. 
  Setting $\delta g_{\mu \nu}|_{\partial} = 0$ is generally impossible, so we need to introduce a counterterm, as disussed above for the case of a scalar field. 
  The precise form of $K = \nabla^\sigma n_\sigma = n_\sigma \left(n^\alpha \nabla_\alpha n^\sigma \right)$ can be inferred from a crude analysis that only Allan understands. 

  What is the physical significance of the boundary term? $K$ measures ``extrinsic'' curvature, roughly speaking how the unit vector $n$ changes as parallel transport it along the boundary. 
  Notice $K$ depends on the way the sub-space is embeded in the total space. 

  


  







  

\end{enumerate}
